Question: Ivy and Andrey were asked to find an explicit formula for the sequence $-100,-50,0,50,...$, where the first term should be $f(1)$. Ivy said the formula is $f(n)=-100+50(n-1)$. Andrey said the formula is $f(n)=-150+50n$. Which one of them is right? Choose 1 answer: Choose 1 answer: (Choice A) A Only Ivy (Choice B) B Only Andrey (Choice C) C Both Ivy and Andrey (Choice D) D Neither Ivy nor Andrey
The general explicit formula for arithmetic sequences is ${a_1}+{d}(n-1)$, where ${a_1}$ is the first term and $ d$ is the common difference. The first term is ${-100}$ and the common difference is ${50}$. ${+50\,\curvearrowright}$ ${+50\,\curvearrowright}$ ${+50\,\curvearrowright}$ ${-100},$ $-50,$ $0,$ $50,...$ We get the following formula. $f(n)={-100}+{50}(n-1)$ So Ivy is definitely right. What about Andrey? We can see that in Andrey's formula, the constant difference is multiplied by $n$ and not by $(n-1)$. Let's expand the parentheses in Ivy's formula to arrive at a similar expression form: $\begin{aligned} f(n)&=-100+50(n-1)\\\\ &=-100+50n-50\\\\ &=-150+50n\end{aligned}$ We obtained Andrey's formula, which means it's also a correct explicit formula for $f(n)$. Both Ivy and Andrey got a correct explicit formula.